Localization at Maximal Ideals Suppose $A$ is a commutative ring with $1\neq0$ satisfying the property that $A_\mathbf{m}$ has no nonzero nilpotent elements for any maximal ideal $\mathbf{m}$, where $$A_\mathbf{m}=S^{-1}A\quad \text{ and }\quad S=A-\mathbf{m}.$$ Prove that $A$ has no nonzero nilpotent elements.
My attempt at the solution was to prove the contrapositive, i.e., suppose $A$ has a nonzero nilpotent element $x$. Then I would like to show that $A_\mathbf{m}$ has a nonzero nilpotent element for some maximal ideal. When I'm using the definition for the equivalence class of fractions, i.e., $$\frac{a}{b}=\frac{a'}{b'}\quad \text{if and only if}\quad (ab'-a'b)u=0$$ for some $u\in S$, I'm having trouble showing that $\frac{x}{s}\neq\frac{0}{1}$. This is because the definition reduces us to $x\cdot u=0$, and it very well seems that $u$ could be the element that makes $x$ a zero divisor. Is there some way to construct an ideal that contains no zero divisors of $x$? My original thought was to take $\mathbf{m}$ to be the maximal ideal containing all of the zero divisors of $x$, but the smallest ideal containing all of the zero divisors very well could be the whole ring (as in the case of $\mathbb{Z}/15\mathbb{Z}$).
Any help or ideas regarding the proof would be very helpful!
 A: Suppose $x$ is a non-zero nilpotent of $A$ like you do. You need a maximal ideal $m$ such that $x/1$ is non-zero in $A_m$. Define the ideal $I = \{s \in A : sx = 0\}$ (show this is an ideal). Now assume that $x/1$ is zero in $A_m$ for all maximal ideals $m$. What can you deduce about the ideal $I$ then? Hence what must $x$ be? Is this a contradiction?
What you actually need to prove, as @Sanchez points out, is that given a ring $A$, if $x \in A$ is nonzero, then $x/1$ must be nonzero in $A_m$ for some maximal ideal $m$.
A: This is more a long comment to Rankeya's answer above than an answer. We recall more generally the following equivalences:


Proposition 3.8 (Atiyah - Macdonald): Let $M$ be an $A$ - module. Then the following are equivalent:




*

*$M = 0$

*$M_\mathfrak{p} = 0$ for all prime ideals $\mathfrak{p}$.

*$M_\mathfrak{m} = 0$ for all maximal ideals $\mathfrak{m}$.


Your problem is then solved by applying the proposition with $M = \mathfrak{R}$ the nilradical of $A$ and noting that the nilradical of $A_\mathfrak{m}$ is in fact $M_\mathfrak{m}$.
